is_scalar() does not consider resource
type values to be scalar as resources are abstract datatypes
which are currently based on integers. This implementation detail should
not be relied upon, as it may change.

Having hunted around the manual, I've not found a clear statement of what makes a type "scalar" (e.g. if some future version of the language introduces a new kind of type, what criterion will decide if it's "scalar"? - that goes beyond just listing what's scalar in the current version.)

A scalar is a single item or value, compared to things like arrays and objects which have multiple values. This tends to be the standard definition of the word in terms of programming. An integer, character, etc are scalars. Strings are probably considered scalars since they only hold "one" value (the value represented by the characters represented) and nothing else.

Hi ... for newbees here, I just want to mention that reference and scalar variable aren't the same. A reference is a pointer to a scalar, just like in C or C++.

<? php // simple reference to scalar

$a = 2; $ref = & $a;

echo "$a <br> $ref";

?>this should print out: "2 <br> 2".

Scalar class also exists. Look below:<? php

class Object_t {

var $a;

function Object_t () // constructor { $this->a = 1; }

}

$a = new Object_t; // we define a scalar object

$ref_a = &a;

echo "$a->a <br> $ref->a";

?>again, this should echo: "1 <br> 1";

Here is another method isued in OOP to acheive on working only over reference to scalar object. Using this, you won't ever have to ask yourself if you work on a copy of the scalar or its reference. You will only possess reference to the scalar object. If you want to duplicate the scalar object, you will have to create a function for that purpose that would read by the reference the values and assign them to another scalar of the same type... or an other type, it is as you wish at that moment.<?php

class objet_t { var $a;

function object_t{$this->a = "patate_poil"; } }

function &get_ref($object_type) {// here we create a scalar object in memory // and we return it by reference to the calling // control scope.return &new $object_type; }

$ref_object_t = get_ref(object_t);

echo "$ref_object_t->a <br>";

?>this should echo: "patate_poit <br>".

The only thing that I try to demonstrate is that scalar variable ARE object in memory while a reference is usualy a variable (scalar object) that contain the address of another scalar object, which contain the informations you want by using the reference.